When running a 100-meter race Ritchie reaches his maximum speed 3 seconds after the race started when he is 13.5 meters from the starting line. Ritchie runs at this max speed for the rest of the race. Ritchie is 48.5 meters from the starting line 8 seconds after the start of the race.




Let's break down the situation with the data provided:

1. **Phase 1: Acceleration phase** (from 0 to 3 seconds)
   - Ritchie reaches his maximum speed 3 seconds after the race starts.
   - In this time, he covers 13.5 meters.

2. **Phase 2: Constant speed phase** (from 3 seconds to the end of the race)
   - After 3 seconds, Ritchie maintains his maximum speed.
   - At 8 seconds after the start, Ritchie has covered 48.5 meters.

### Step 1: Calculating the acceleration phase
The acceleration phase lasts 3 seconds, and in this time, Ritchie covers 13.5 meters. Assuming constant acceleration, we can use the equation for distance traveled under constant acceleration:

\[
d = \frac{1}{2} a t^2
\]

Here:
- \( d = 13.5 \, \text{meters} \) (distance covered in 3 seconds)
- \( t = 3 \, \text{seconds} \)
- \( a \) is the acceleration.

Solving for \( a \):

\[
13.5 = \frac{1}{2} a (3)^2
\]
\[
13.5 = \frac{1}{2} a (9)
\]
\[
13.5 = 4.5 a
\]
\[
a = 3 \, \text{m/s}^2
\]

So, Ritchie's acceleration during the first 3 seconds is \( 3 \, \text{m/s}^2 \).

### Step 2: Calculating Ritchie's maximum speed
Ritchie reaches his maximum speed at the end of the acceleration phase, which is after 3 seconds. We can calculate the maximum speed using the formula:

\[
v = a t
\]

Where:
- \( a = 3 \, \text{m/s}^2 \)
- \( t = 3 \, \text{seconds} \)

\[
v = 3 \times 3 = 9 \, \text{m/s}
\]

So, Ritchie's maximum speed is \( 9 \, \text{m/s} \).

### Step 3: Checking the distance at 8 seconds
Now we need to check how far Ritchie has traveled after 8 seconds.

- In the first 3 seconds, he covers 13.5 meters.
- In the next 5 seconds (from 3 to 8 seconds), he runs at his maximum speed of \( 9 \, \text{m/s} \).

The distance covered in 5 seconds at constant speed is:

\[
d = v t = 9 \times 5 = 45 \, \text{meters}
\]

So, the total distance after 8 seconds is:

\[
13.5 + 45 = 58.5 \, \text{meters}
\]

But the problem states that after 8 seconds, Ritchie has only covered **48.5 meters**. There seems to be a discrepancy here. Would you like to adjust or clarify the information, or explore further possibilities, such as different interpretations of the data?

---

Let me know if you'd like more details on any part of the explanation or have any questions! Here are 5 related questions you might explore:

1. What is the role of constant acceleration in determining speed?
2. How can you derive time from distance and speed during constant motion?
3. How do acceleration and speed interact during different phases of a race?
4. Could deceleration ever be a factor in a sprint like this?
5. How might environmental factors like wind or terrain affect Ritchie's speed?

**Tip:** When calculating distances during different phases of motion, always separate acceleration and constant speed phases for clarity.

When running a 100-meter race, Ritchie reaches his maximum speed 3 seconds after the race started when he is 13.5 meters from the starting line. Ritchie runs at this max speed for the rest of the race. Ritchie is 48.5 meters from the starting line 8 seconds after the start of the race.

Explore a detailed solution to a kinematics problem where Ritchie runs a 100-meter race. Learn how to calculate acceleration, speed, and distance using the equations of motion under constant acceleration. Suitable for Grades 9-12 students.

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